Circle

Exercise 11.3

Question 7: Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution: Given, diameter of the semicircle = 10cm.

Circumference of the circle = π D

= 3.14 xx 10 = 31.4

So, perimeter of semicircle = 31.4 ÷ 2 = 15.7 cm

Perimeter of the given figure = 15.7 + 10 = 25.7 cm

Question 8: Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m^2. (Take π = 3.14)

Solution: To calculate the cost of polishing of circular table, we have to calculate its area first of all.Here, Given diameter of the circular table-top = 1.6 m. Hence, r = 0.8 m

Area = π r^2

= 3.14 xx 0.8 xx 0.8 = 2.0096 sq m

Cost of polishing = Rate xx Area

= Rs. 15 xx 2.0096 = Rs. 30.144

Question 9: Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?

Solution: The length of the wire = 44 cm = circumference of the circle, Radius = ?

When the same wire is bent into the shape of square,

Perimeter of the square = length of the wire = 44 cm

To know that which shape will encloses more area, we have to calculate their areas.

Circumference = 2 π r

Or, 44 = 2 xx (22)/(7) xx\ r

Or, r = 44 xx (7)/(44) = 7 cm

Now, Area = π r^2

= (22)/(7) xx 7 xx 7 = 22 xx 7 = 154 sq cm

When the same wire is bent to the shape of a square

perimeter of a square = 4 xx \text(side)

Or, 44 = 4 xx \text(side)

Or, side = 44 ÷ 4 = 11 cm

Area of a square = side2= 11^2= 121 cm2

It is clear; area of the circle > Area of square

Question 10: From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet.

Solution: Radius of circular card sheet = 14 cm

Radius of each circle which has been removed = 3.5 cm

Length of the rectangle which has been removed = 3cm

Breadth of the rectangle which has been removed = 1 cm

Area of circle = π r^2

= (22)/(7) xx 14 xx 14

= 22 xx 2 xx 14 = 616 sq cm

Now,Area of the circle which is removed = π r^2

= (22)/(7) xx 7/2 xx 7/2

= 11 xx 7/2 = 38.5 sq cm

Area of two such circles = 2 xx 38.5= 77 cm2

Area of the rectangle which has been removed = length xx breadth

= 3xx1= 3 cm2

Total area of the shapes which has been removed = Area of two circles + area of rectangle

= 77cm2+ 3 cm2= 80 cm2

Area left after removing the two circles and one rectangle

= Area of circular sheet — Total area of the shapes which has been removed

= 616 cm2 — 80 cm2 = 536 sq cm